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Question

A circuit has a \( 10 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and two
resistors \( 4 \, \Omega \) and \( 6 \, \Omega \) in parallel. What is the total current?

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Explanation

**Series combination** R_eq = R₁+R₂+..., same current I through each, voltage divides proportionally V_i = I R_i. Parallel combination 1/R_p = 1/R₁+1/R₂+..., same voltage V across each, current divides inversely, equivalent R_p = (R₁ R₂)/(R₁+R₂) for two resistors. Parallel resistance: (1/R_p) = (1/4) + (1/6) = (3 + 2/12) = (5/12) ⇒ R_p = (12/5) = 2.4 Ω . Total resistance: Rtₒtₐl = 2 + 2.4 = 4.4 Ω . Current: I = (ε/Rtₒtₐl) = (10/4.4) ≈ 2.27 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r

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